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Your task is to draw lines between edges on a regular pentagon such that if you tile a dodecahedron with 12 identical copies of that pentagon you get a single closed line which does not intersect itself and which visits each face of the dodecahedron exactly twice. How many essentially different solutions exist?

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    $\begingroup$ Is it allowed to run a line along the edge of a pentagon such that it may go between two pentagons that are non-adjacent, and is not considered to 'visit' either of the other two pentagons that the edge divides? $\endgroup$ Commented Sep 11, 2023 at 10:47
  • $\begingroup$ Can copies of the pentagonal tile be reflected? $\endgroup$ Commented Sep 11, 2023 at 14:23
  • $\begingroup$ @Weather Vane: I do not exactly understand what you mean with a line that goes between two pentagons that are not adjacent. With visiting the face of the dodecahedron I mean that the line must cross some edge of that face such that the line has points with the inner of this face in common. $\endgroup$ Commented Sep 11, 2023 at 16:47
  • $\begingroup$ @AxiomaticSystem: No, the pentagons must be really identical. $\endgroup$ Commented Sep 11, 2023 at 16:48
  • $\begingroup$ If the line crosses the edge at a vertex then it need not cross the edge of the adjoining pentagon, but could follow another edge, to another polygon that the first one does not share an edge with, entering again at a vertex. $\endgroup$ Commented Sep 11, 2023 at 16:54

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2nd Update: I have found a small number of solutions.

I found 6 distinct solutions with a continuous path.

For a line to visit each face twice, there must be 2 entry and 2 exit points.
There are 10 distinct ways to arrange them around a pentagon.
These exclude reflections and rotations.
Moreover, there are two ways for the points to connect without crossing.
Here are the ten face arangements, with the joining variation in vertical pairs.

enter image description here

Only 4 of those 10 sets will make a dodecahedron with matching edges.
And only one (top right) will make a continuous path.
(It's clear on inspection that those with a closed edge loop will not.)

There were 210 possible arrangements.
Only 6 after removing rotational and reflection symmetry.
There are 3 versions of each of the path methods.

enter image description here

enter image description here

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  • $\begingroup$ The solutions are very "samey" within the distinct basic types described. With that many images I can't yet figure out how to compare them – it's impractical to do it empirically. Algorithmically I would have to align every face of the candidate with every face of every polyhedron in the 'solutions' list and check if they are in same orientation (and the mirror equivalents). $\endgroup$ Commented Sep 12, 2023 at 17:59
  • $\begingroup$ @ Weather Vane: Nice result so far! For the tiling with the pentagon with only small curves the number of essentially different tilings can be easily derived manually with the result from puzzling.stackexchange.com/questions/121943/…. For the mixed case this should work too. $\endgroup$ Commented Sep 12, 2023 at 18:08
  • $\begingroup$ Thanks, I am unconvinced that there are no 'continuous path' solutions for the other basic types. After finding that only some of the basic types could be built into a dodecahedron, at first I could not find any solutions with a continuous path, and it's possible that solutions for other types are hiding behind 'deficient code'. $\endgroup$ Commented Sep 12, 2023 at 18:14
  • $\begingroup$ @Herbert I've done rotational symmetry but I can't get my head around mirror symmetry yet. $\endgroup$ Commented Sep 13, 2023 at 13:39
  • $\begingroup$ @ Weather Vane First I agree that the only possible types are the two top right cases. Concerning the essential different types I really recommend to look just at the 5 ways the double tiles can be placed in each of both cases and to check if these make up a single closed loop. $\endgroup$ Commented Sep 13, 2023 at 14:22

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